Problem: Solve for $x$ and $y$ using elimination. $\begin{align*}3x+7y &= -3 \\ 5x+5y &= 5\end{align*}$
Answer: We can eliminate $x$ when its corresponding coefficients are negative inverses. Recalling our knowledge of least common multiples, multiply the top equation by $-5$ and the bottom equation by $3$ $\begin{align*}-15x-35y &= 15\\ 15x+15y &= 15\end{align*}$ Add the top and bottom equations. $-20y = 30$ Divide both sides by $-20$ and reduce as necessary. $y = -\dfrac{3}{2}$ Substitute $-\dfrac{3}{2}$ for $y$ in the top equation. $3x+7( -\dfrac{3}{2}) = -3$ $3x-\dfrac{21}{2} = -3$ $3x = \dfrac{15}{2}$ $x = \dfrac{5}{2}$ The solution is $\enspace x = \dfrac{5}{2}, \enspace y = -\dfrac{3}{2}$.